Limit of $2^{(\log_2 n)^2}/2^{(\log_2 n)^3}$

Question

I am trying to find the following limit

$$\lim_{n \to \infty} \frac{2^{(\log_2 n)^2}} {2^{(\log_2 n)^3}}$$

I really don't know where to start and any help would be appreciated!

Answer 1

Hint

What is the limit of the logarithm of your expression?

Answer 2

We're going to apply two of the Laws of Logs. We will use

\begin{eqnarray*} \log_b(x^y) &=& y\log_b(x) \\ \\ \log_b\!\left(\frac{x}{y}\right) &=& \log_b(x)-\log_b(y) \end{eqnarray*}

I'm going to take the log of your expression:

\begin{eqnarray*} \log_2\left(\frac{2^{(\log_2 n)^2}}{2^{(\log_2 n)^3}}\right) &=& \log_2\left(2^{(\log_2 n)^2}\right) - \log_2\left(2^{(\log_2 n)^3}\right) \\ \\ &=& (\log_2 n)^2\cdot\log_2(2) - (\log_2 n)^3\cdot\log_2(2) \\ \\ &=& (\log_2 n)^2 - (\log_2 n)^3 \\ \\ &=& (\log_2 n)^2 \cdot (1-\log_2n) \end{eqnarray*}

Hopefully, you agree that $\log_2n \to \infty$ as $n \to \infty$. This means that $$\lim_{n \to \infty}(\log_2 n)^2 \cdot (1-\log_2n) = -\infty$$

Going back a few lines, tells us then that $$\lim_{n \to \infty} \log_2\left(\frac{2^{(\log_2 n)^2}}{2^{(\log_2 n)^3}}\right) = -\infty$$

Looking at the graph $y = \log_2 x$ shows us that $\log_2 x \to -\infty$ as $x \to 0$ and hence $$\lim_{n \to \infty} \frac{2^{(\log_2 n)^2}}{2^{(\log_2 n)^3}} = 0$$

Answer 3

Hint: $$\frac{2^{\log_2(n)^2}}{2^{\log_2(n)^3}}=2^{\log_2(n)^2-\log_2(n)^3}$$ What is $\lim\limits_{n\to\infty}\left(\log_2(n)^2-\log_2(n)^3\right)$?

Answer 4

Look at it this way:

$\log_2 n$ gets large for large $n$. Therefore, $(\log_2 n)^3$ gets larger than $(\log_2 n)^2$.

Therefore $2^{(\log_2 n)^3}$ gets much larger than $2^{(\log_2 n)^2}$.

Therefore $\frac{2^{(\log_2 n)^2}}{2^{(\log_2 n)^3}} \to 0 $.

The rest is details.

Answer 5

After cancelling the numerator with the denominator what we get is 1/2^(log n to the base 2) After that use the fact that X ^(log Y to the base X) is nothing but equal to Y itself. So in your case it would simply be 1/n And now if n is tending to infinity and is surely 0...